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1. 1.Computation · 7 min

   Let $S$ be the set of points in ${\mathbb{R}}^3$ satisfying

   $$x^2+y^2+z^2-2x+6y-4z+5=0.$$

   1. (a)Show that $S$ is a sphere, and find its centre $C$ and radius $r$.
   2. (b)Decide, without any decimal approximation, whether the origin lies inside, on, or outside $S$. Give two independent justifications.
   3. (c)For which values of $k$ does the plane $z=k$ meet $S$? Describe the intersection of $S$ with the plane $z=4$ exactly (what curve, its centre, its radius).
   4. (d)Describe in words the region $x^2+y^2+z^2-2x+6y-4z+5\le 0,y\ge -3$.
   Show the solution to question 1

   (a) Complete the square in each variable. Group the terms by variable and move the constant across:

   $$(x^2-2x)+(y^2+6y)+(z^2-4z)=-5.$$

   Add $1$, $9$ and $4$ to both sides to complete the three squares:

   $$(x-1{)}^2+(y+3{)}^2+(z-2{)}^2=-5+1+9+4=9.$$

   This is the equation of the sphere with centre $C(1,-3,2)$ and radius $r=\sqrt{9}=3$. The right-hand side came out positive, so it is a genuine sphere and not a point or the empty set.

   (b) Compare a distance with the radius. By the distance formula, $|OC|=\sqrt{1^2+(-3{)}^2+2^2}=\sqrt{14}$. Since $14>9$, $\sqrt{14}>3=r$, so the origin is farther from the centre than the radius: it lies outside $S$.

   Second justification: the original left-hand side equals $(x-1{)}^2+(y+3{)}^2+(z-2{)}^2-9=|PC{|}^2-9$, which is negative inside the sphere, zero on it and positive outside. At the origin it equals $0+0+0+5=5>0$, so outside. (Equivalently $|OC{|}^2=14=9+5$.)

   (c) Slice with a horizontal plane. Substituting $z=k$ into the completed-square form gives

   $$(x-1{)}^2+(y+3{)}^2=9-(k-2{)}^2.$$

   This has solutions exactly when $9-(k-2{)}^2\ge 0$, that is $|k-2|\le 3$, so $-1\le k\le 5$. At the two extremes $k=-1$ and $k=5$ the plane is tangent and the intersection is a single point; at $k=2$ it passes through the centre and cuts a great circle of radius $3$.

   For $k=4$ the equation is $(x-1{)}^2+(y+3{)}^2=9-4=5$: a circle lying in the plane $z=4$, with centre $(1,-3,4)$ (directly above $C$) and radius $\sqrt{5}$.

   (d) Read the inequalities geometrically. The first inequality is $|PC{|}^2\le 9$, the solid ball of radius $3$ centred at $C(1,-3,2)$. The plane $y=-3$ passes through the centre, and $y\ge -3$ keeps the half of the ball on one side of it. The region is a solid hemisphere (half-ball) of radius $3$: its flat face is the disc of radius $3$ centred at $C$ in the plane $y=-3$, and its curved face bulges toward positive $y$.

2. 2.Computation · 6 min

   Let $\mathbf{a}=⟨2,-1,2⟩$ and $\mathbf{b}=⟨1,2,-2⟩$.

   1. (a)Find $|\mathbf{a}|$, the unit vector in the direction of $\mathbf{a}$, and the vector of length $6$ pointing in the direction opposite to $\mathbf{a}$.
   2. (b)Find the vector $\mathbf{c}$ satisfying $\mathbf{a}+2\mathbf{c}=\mathbf{b}$, and check your answer.
   3. (c)Compute $|\mathbf{a}+\mathbf{b}|$ and $|\mathbf{a}|+|\mathbf{b}|$ exactly. Which is larger, which inequality says this must always be so, and what would have to be true of $\mathbf{a}$ and $\mathbf{b}$ for the two to be equal?
   4. (d)Find the point $Q$ such that $\overrightarrow{PQ}=2\mathbf{a}-\mathbf{b}$, where $P=(3,0,-1)$.
   Show the solution to question 2

   (a) Normalise. $|\mathbf{a}|=\sqrt{4+1+4}=\sqrt{9}=3$. The unit vector in the direction of $\mathbf{a}$ is

   $$\mathbf{u}=\frac{\mathbf{a}}{|\mathbf{a}|}=⟨\frac{2}{3},-\frac{1}{3},\frac{2}{3}⟩.$$

   A vector of length $6$ in the opposite direction is $-6\mathbf{u}=⟨-4,2,-4⟩$. Check: $\sqrt{16+4+16}=\sqrt{36}=6$, and it is a negative multiple of $\mathbf{a}$, so it points the opposite way.

   (b) Solve for the unknown vector exactly as for a number. From $\mathbf{a}+2\mathbf{c}=\mathbf{b}$, $\mathbf{c}=\frac{1}{2}(\mathbf{b}-\mathbf{a})=\frac{1}{2}⟨1-2,2+1,-2-2⟩=\frac{1}{2}⟨-1,3,-4⟩=⟨-\frac{1}{2},\frac{3}{2},-2⟩$.

   Check: $\mathbf{a}+2\mathbf{c}=⟨2-1,-1+3,2-4⟩=⟨1,2,-2⟩=\mathbf{b}$.

   (c) Magnitude is not additive. $\mathbf{a}+\mathbf{b}=⟨3,1,0⟩$, so $|\mathbf{a}+\mathbf{b}|=\sqrt{9+1+0}=\sqrt{10}$. Meanwhile $|\mathbf{b}|=\sqrt{1+4+4}=3$, so $|\mathbf{a}|+|\mathbf{b}|=6$. Since $10<36$, $\sqrt{10}<6$: the sum of the lengths is larger.

   This is the triangle inequality $|\mathbf{a}+\mathbf{b}|\le |\mathbf{a}|+|\mathbf{b}|$, with equality only when the vectors point the same way, i.e. one is a positive scalar multiple of the other. Here $\mathbf{a}$ and $\mathbf{b}$ are not parallel at all (the ratio of first components is $2$, of second components $-\frac{1}{2}$), so strict inequality was guaranteed before any arithmetic.

   (d) Head equals tail plus displacement. $2\mathbf{a}-\mathbf{b}=⟨4,-2,4⟩-⟨1,2,-2⟩=⟨3,-4,6⟩$. Since $\overrightarrow{PQ}=Q-P$, $Q=P+⟨3,-4,6⟩=(3+3,0-4,-1+6)=(6,-4,5)$.

3. 3.Proof · 8 min

   Throughout, use only the algebraic properties of the dot product ($\mathbf{a}\cdot \mathbf{a}=|\mathbf{a}{|}^2$, commutativity, distributivity, scalars pull out); components may not be used.

   1. (a)Prove the parallelogram identity $|\mathbf{a}+\mathbf{b}{|}^2+|\mathbf{a}-\mathbf{b}{|}^2=2|\mathbf{a}{|}^2+2|\mathbf{b}{|}^2$, and say what it asserts about the sides and diagonals of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$.
   2. (b)Prove that $(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=|\mathbf{a}{|}^2-|\mathbf{b}{|}^2$, and deduce that the diagonals of a parallelogram are perpendicular if and only if all four sides have the same length.
   3. (c)Two vectors satisfy $|\mathbf{a}|=3$, $|\mathbf{b}|=5$ and $|\mathbf{a}+\mathbf{b}|=7$. Find $|\mathbf{a}-\mathbf{b}|$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ exactly.
   Show the solution to question 3

   (a) Expand both squares with the dot product. Using $|\mathbf{v}{|}^2=\mathbf{v}\cdot \mathbf{v}$, distributivity and commutativity,

   $$|\mathbf{a}+\mathbf{b}{|}^2=(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}+\mathbf{b})=\mathbf{a}\cdot \mathbf{a}+2\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{b}=|\mathbf{a}{|}^2+2\mathbf{a}\cdot \mathbf{b}+|\mathbf{b}{|}^2,$$

   and in the same way $|\mathbf{a}-\mathbf{b}{|}^2=|\mathbf{a}{|}^2-2\mathbf{a}\cdot \mathbf{b}+|\mathbf{b}{|}^2$. Adding, the cross terms cancel and

   $$|\mathbf{a}+\mathbf{b}{|}^2+|\mathbf{a}-\mathbf{b}{|}^2=2|\mathbf{a}{|}^2+2|\mathbf{b}{|}^2.■$$

   Geometrically, $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ are the two diagonals of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$, so the identity says: the sum of the squares of the two diagonals equals the sum of the squares of the four sides.

   (b) Subtract the same two expansions instead of adding. Directly,

   $$(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\mathbf{a}\cdot \mathbf{a}-\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{a}-\mathbf{b}\cdot \mathbf{b}=|\mathbf{a}{|}^2-|\mathbf{b}{|}^2,$$

   because $\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$. $■$

   For a genuine parallelogram $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, so both diagonals are nonzero vectors. Nonzero vectors are perpendicular if and only if their dot product is zero, and by the identity that happens if and only if $|\mathbf{a}{|}^2=|\mathbf{b}{|}^2$, i.e. $|\mathbf{a}|=|\mathbf{b}|$. Opposite sides of a parallelogram are equal anyway, so $|\mathbf{a}|=|\mathbf{b}|$ says all four sides are equal: the diagonals are perpendicular exactly when the parallelogram is a rhombus.

   (c) Feed the data into (a) and into the expansion. From the parallelogram identity,

   $$|\mathbf{a}-\mathbf{b}{|}^2=2\cdot 9+2\cdot 25-49=68-49=19,|\mathbf{a}-\mathbf{b}|=\sqrt{19}.$$

   For the angle, the expansion of $|\mathbf{a}+\mathbf{b}{|}^2$ gives $49=9+2\mathbf{a}\cdot \mathbf{b}+25$, so $\mathbf{a}\cdot \mathbf{b}=\frac{15}{2}$. By the geometric form of the dot product,

   $$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{15/2}{15}=\frac{1}{2},\theta =\frac{\pi }{3}.$$

   Cross-check with the other expansion: $|\mathbf{a}-\mathbf{b}{|}^2=9+25-2\cdot \frac{15}{2}=19$, agreeing with the first calculation.

4. 4.Computation · 7 min

   Let $\mathbf{a}=⟨2,1,-1⟩$, $\mathbf{b}=⟨1,-1,3⟩$ and $\mathbf{c}=⟨0,3,-7⟩$.

   1. (a)Compute $\mathbf{a}\times \mathbf{b}$ from the determinant, verify that it is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, and write down both unit vectors perpendicular to $\mathbf{a}$ and $\mathbf{b}$.
   2. (b)Find the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$, and the area of the triangle with vertices $O$, $O+\mathbf{a}$ and $O+\mathbf{b}$.
   3. (c)Compute the scalar triple product $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ two different ways. State what it tells you about the parallelepiped with edges $\mathbf{a},\mathbf{b},\mathbf{c}$, and then express $\mathbf{c}$ as a combination $s\mathbf{a}+t\mathbf{b}$.
   Show the solution to question 4

   (a) Expand along the first row.

   $$\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 1& -1\\ 1& -1& 3\end{array}\right|=(1\cdot 3-(-1)(-1))\mathbf{i}-(2\cdot 3-(-1)\cdot 1)\mathbf{j}+(2\cdot (-1)-1\cdot 1)\mathbf{k}=⟨2,-7,-3⟩.$$

   The middle component carries the minus sign: $-(6+1)=-7$. Check perpendicularity: $⟨2,-7,-3⟩\cdot \mathbf{a}=4-7+3=0$ and $⟨2,-7,-3⟩\cdot \mathbf{b}=2+7-9=0$.

   $|\mathbf{a}\times \mathbf{b}|=\sqrt{4+49+9}=\sqrt{62}$, so the two unit vectors perpendicular to both are $\pm \frac{1}{\sqrt{62}}⟨2,-7,-3⟩$. There are exactly two because the perpendicular direction to a plane is unique up to sign.

   (b) Area is the length of the cross product. The parallelogram has area $|\mathbf{a}\times \mathbf{b}|=\sqrt{62}$. The triangle with those two sides from the common vertex is half of it: $\frac{\sqrt{62}}{2}$.

   (c) First way: use the product already computed. Since $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}$,

   $$(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}=⟨2,-7,-3⟩\cdot ⟨0,3,-7⟩=0-21+21=0.$$

   Second way: the determinant. Expanding along the first row,

   $$\left|\begin{array}{ccc}2& 1& -1\\ 1& -1& 3\\ 0& 3& -7\end{array}\right|=2\left|\begin{array}{cc}-1& 3\\ 3& -7\end{array}\right|-1\left|\begin{array}{cc}1& 3\\ 0& -7\end{array}\right|+(-1)\left|\begin{array}{cc}1& -1\\ 0& 3\end{array}\right|=2(7-9)-(-7-0)-(3-0)=-4+7-3=0.$$

   (Or compute $\mathbf{b}\times \mathbf{c}=⟨7-9,-(-7-0),3-0⟩=⟨-2,7,3⟩$ and dot with $\mathbf{a}$: $-4+7-3=0$.)

   The parallelepiped has volume $|0|=0$: it is flat, so the three vectors are coplanar. Being coplanar, and with $\mathbf{a}$, $\mathbf{b}$ not parallel, $\mathbf{c}$ must be a combination of the other two. Solve $\mathbf{c}=s\mathbf{a}+t\mathbf{b}$ componentwise:

   $$2s+t=0,s-t=3,-s+3t=-7.$$

   Adding the first two gives $3s=3$, so $s=1$ and $t=-2$. The third equation is the check: $-1-6=-7$. Hence $\mathbf{c}=\mathbf{a}-2\mathbf{b}$; verify: $⟨2,1,-1⟩-⟨2,-2,6⟩=⟨0,3,-7⟩$.

5. 5.Computation · 9 min

   Consider the lines

   $$L_1:x=1+t,y=2+t,z=2t\text{and}{L}_2:x=2s,y=1-s,z=3+s.$$

   1. (a)Determine whether $L_1$ and $L_2$ are parallel, intersecting or skew. Show the full test, including the equation that fails.
   2. (b)Find the distance between the two lines exactly.
   3. (c)Find an equation of the plane that contains $L_1$ and is parallel to $L_2$. Then use that plane to confirm your answer to (b) by a second route.
   4. (d)Find the point where $L_1$ meets the plane $2x-y+z=7$, and check it.
   Show the solution to question 5

   (a) Compare directions, then try to solve for a common point. The direction vectors are ${\mathbf{v}}_1=⟨1,1,2⟩$ and ${\mathbf{v}}_2=⟨2,-1,1⟩$. Neither is a scalar multiple of the other (the first components would need the factor $2$, the second components $-1$), so the lines are not parallel.

   If they meet, some $t$ and $s$ satisfy all three coordinate equations:

   $$1+t=2s,2+t=1-s,2t=3+s.$$

   Subtracting the first equation from the second eliminates $t$: $1=1-3s$, so $s=0$, and then $t=2s-1=-1$. Test these values in the third equation: the left side is $2t=-2$, the right side is $3+s=3$. They disagree, so no common point exists. Not parallel and not intersecting: the lines are skew.

   (b) Distance via a common normal. The lines lie in a pair of parallel planes whose common normal is perpendicular to both directions:

   $$\mathbf{n}={\mathbf{v}}_1\times {\mathbf{v}}_2=⟨1\cdot 1-2\cdot (-1),-(1\cdot 1-2\cdot 2),1\cdot (-1)-1\cdot 2⟩=⟨3,3,-3⟩,$$

   and any nonzero multiple will do, so take $\mathbf{n}=⟨1,1,-1⟩$. (Check: $\mathbf{n}\cdot {\mathbf{v}}_1=1+1-2=0$, $\mathbf{n}\cdot {\mathbf{v}}_2=2-1-1=0$.) Pick $P_1(1,2,0)$ on $L_1$ ($t=0$) and $P_2(0,1,3)$ on $L_2$ ($s=0$), so $\overrightarrow{P_1{P}_2}=⟨-1,-1,3⟩$. The distance is the absolute scalar projection of this vector onto $\mathbf{n}$:

   $$D=\frac{|\mathbf{n}\cdot \overrightarrow{P_1{P}_2}|}{|\mathbf{n}|}=\frac{|-1-1-3|}{\sqrt{3}}=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}.$$

   (c) A plane through $L_1$ with normal $\mathbf{n}$. A plane containing $L_1$ and parallel to $L_2$ must have a normal perpendicular to both ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$, which is exactly $\mathbf{n}$. Through $P_1(1,2,0)$:

   $$1(x-1)+1(y-2)-1(z-0)=0⟺x+y-z=3.$$

   It contains $L_1$: substituting gives $(1+t)+(2+t)-2t=3$ for every $t$. It is parallel to $L_2$: substituting gives $2s+(1-s)-(3+s)=-2$ for every $s$, a constant, so $L_2$ never crosses the plane and in fact lies in the parallel plane $x+y-z=-2$.

   Second route to (b): the distance from any point of $L_2$, say $(0,1,3)$, to the plane $x+y-z-3=0$ is

   $$\frac{|0+1-3-3|}{\sqrt{1+1+1}}=\frac{5}{\sqrt{3}},$$

   the same value. (Equivalently the two parallel planes $x+y-z=3$ and $x+y-z=-2$ are $|3-(-2)|/\sqrt{3}=5/\sqrt{3}$ apart.)

   (d) Substitute the parametric equations into the plane.

   $$2(1+t)-(2+t)+2t=7⟹2+2t-2-t+2t=7⟹3t=7,t=\frac{7}{3}.$$

   The point is $\left(1+\frac{7}{3},2+\frac{7}{3},\frac{14}{3}\right)=\left(\frac{10}{3},\frac{13}{3},\frac{14}{3}\right)$. Check in the plane: $\frac{20}{3}-\frac{13}{3}+\frac{14}{3}=\frac{21}{3}=7$. (A single solution was guaranteed because ${\mathbf{v}}_1\cdot ⟨2,-1,1⟩=2-1+2=3\ne 0$: the line is not parallel to the plane.)

6. 6.Sketch · 10 min

   Consider the surface

   $$4x^2+y^2-z^2-8x+2y+1=0.$$

   A sketch with unlabelled curves earns nothing here: every trace you draw must be named, with its centre and dimensions.

   1. (a)Complete the square and bring the equation to a standard form. Identify the surface, its centre, and the direction of its axis.
   2. (b)Find and describe the traces in the planes $z=0$, $z=\pm 2$, $x=1$ and $y=-1$ — name each curve and give its key dimensions (semi-axes, vertices, asymptotes).
   3. (c)Sketch the surface, showing the traces from (b), the centre, and the axis. Say which way the surface opens and why it is connected.
   4. (d)Suppose the constant term $1$ is replaced by a constant $k$. For which $k$ is the surface a cone? A hyperboloid of two sheets? Justify from the standard form.
   Show the solution to question 6

   (a) Complete the square in $x$ and in $y$. Group and factor the leading coefficient out of the $x$-terms:

   $$4(x^2-2x)+(y^2+2y)-z^2+1=0⟹4[(x-1{)}^2-1]+[(y+1{)}^2-1]-z^2+1=0.$$

   Collecting constants, $4(x-1{)}^2+(y+1{)}^2-z^2=4$, and dividing by $4$,

   $$(x-1{)}^2+\frac{(y+1{)}^2}{4}-\frac{z^2}{4}=1.$$

   Three squared terms, exactly one negative, and $1$ on the right: a hyperboloid of one sheet. Its centre is $(1,-1,0)$ and its axis is the line $x=1$, $y=-1$, parallel to the $z$-axis — the variable carrying the minus sign.

   (b) Slice. Substitute the constant into the standard form.

   $z=0$: $(x-1{)}^2+\frac{(y+1{)}^2}{4}=1$, an ellipse centred at $(1,-1,0)$ with semi-axes $1$ in the $x$-direction and $2$ in the $y$-direction. This is the waist, the smallest horizontal cross-section.

   $z=\pm 2$: $(x-1{)}^2+\frac{(y+1{)}^2}{4}=1+1=2$, ellipses centred at $(1,-1,\pm 2)$ with semi-axes $\sqrt{2}$ and $2\sqrt{2}$ — the same shape as the waist, scaled by $\sqrt{2}$.

   $x=1$: $\frac{(y+1{)}^2}{4}-\frac{z^2}{4}=1$, i.e. $(y+1{)}^2-z^2=4$: a hyperbola in the plane $x=1$ opening in the $\pm y$ directions, vertices at $(1,1,0)$ and $(1,-3,0)$, asymptotes $z=\pm (y+1)$.

   $y=-1$: $(x-1{)}^2-\frac{z^2}{4}=1$: a hyperbola in the plane $y=-1$ opening in the $\pm x$ directions, vertices at $(2,-1,0)$ and $(0,-1,0)$, asymptotes $z=\pm 2(x-1)$.

   (c) The drawing. Draw the axis as a vertical line through $(1,-1,0)$ and mark the centre. Draw the waist ellipse at height $0$ (width $2$ in $x$, $4$ in $y$), and the larger ellipses at heights $2$ and $-2$ (width $2\sqrt{2}$ in $x$, $4\sqrt{2}$ in $y$). Join them with the two hyperbolic profiles: in the front-to-back plane $y=-1$ the outline passes through $(0,-1,0)$ and $(2,-1,0)$ and flares outward above and below; in the side plane $x=1$ it passes through $(1,-3,0)$ and $(1,1,0)$ and flares likewise. The surface opens along the $z$-axis in both directions, getting wider without bound as $|z|$ grows (the ellipses at height $z$ have semi-axes $\sqrt{1+z^2/4}$ and $2\sqrt{1+z^2/4}$). It is connected because every horizontal plane meets it in a nonempty ellipse: $1+z^2/4>0$ for all $z$, so there is no gap.

   (d) Vary the constant. With constant $k$ the completed-square computation gives

   $$4(x-1{)}^2+(y+1{)}^2-z^2=5-k.$$

   If $k=5$ the right-hand side is $0$: $z^2=4(x-1{)}^2+(y+1{)}^2$ is an elliptic cone with vertex $(1,-1,0)$, and the horizontal traces pinch to a point there. If $k>5$ the right-hand side is negative; dividing by it flips the signs to $-\frac{4(x-1{)}^2}{k-5}-\frac{(y+1{)}^2}{k-5}+\frac{z^2}{k-5}=1$, two negative squared terms: a hyperboloid of two sheets, with no points where $z^2<k-5$. If $k<5$ the right-hand side is positive and the surface is a hyperboloid of one sheet, as in (a). The cone at $k=5$ is the boundary case between the two hyperboloids, and it is their common asymptotic surface.

7. 7.Computation · 9 min

   The paraboloid $z=x^2+y^2$ and the plane $z=2y$ meet in a closed curve $C$.

   1. (a)First a warm-up on a different function: find the domain of $\mathbf{r}(t)=⟨\ln (t+1),\sqrt{4-t^2},\frac{1}{t-1}⟩$, and evaluate $\underset{t\to 1}{\lim }⟨\frac{t^2-1}{t-1},\ln t,\sqrt{t+3}⟩$.
   2. (b)Find a vector function that parametrises $C$ once, and verify that every point of your curve lies on both surfaces. Then find the highest and lowest points of $C$.
   3. (c)Find all points where $C$ meets the plane $x=y$. If your method introduces extraneous solutions, identify and discard them explicitly.
   Show the solution to question 7

   (a) Intersect the component domains; take the limit componentwise. $\ln (t+1)$ needs $t>-1$; $\sqrt{4-t^2}$ needs $-2\le t\le 2$; $\frac{1}{t-1}$ needs $t\ne 1$. The domain is the intersection, $(-1,1)\cup (1,2]$.

   For the limit, each component is its own single-variable problem. $\frac{t^2-1}{t-1}=t+1$ for $t\ne 1$, so the first component tends to $2$; $\ln t\to \ln 1=0$ by continuity; $\sqrt{t+3}\to \sqrt{4}=2$ by continuity. The limit is $⟨2,0,2⟩$.

   (b) Project onto the $xy$-plane, then lift. A point on both surfaces satisfies $x^2+y^2=2y$, that is $x^2+(y-1{)}^2=1$ after completing the square. So the projection of $C$ onto the $xy$-plane is the circle of radius $1$ centred at $(0,1)$. Parametrise it by $x=\cos t$, $y=1+\sin t$, and let the plane supply the height $z=2y$:

   $$\mathbf{r}(t)=⟨\cos t,1+\sin t,2+2\sin t⟩,0\le t\le 2\pi .$$

   Verification on the paraboloid: $x^2+y^2={\cos }^{2}t+1+2\sin t+{\sin }^{2}t=2+2\sin t=z$. On the plane: $2y=2+2\sin t=z$. Both hold for every $t$, and as $t$ runs over $[0,2\pi )$ the projection goes once round its circle, so $C$ is traversed once. (Since $C$ lies in the plane $z=2y$ and projects to a circle, it is an ellipse.)

   The height is $z=2+2\sin t$, largest when $\sin t=1$ and smallest when $\sin t=-1$. Highest point: $t=\frac{\pi }{2}$, $\mathbf{r}=(0,2,4)$. Lowest point: $t=\frac{3\pi }{2}$, $\mathbf{r}=(0,0,0)$. Check: $(0,2,4)$ has $0+4=4=2\cdot 2$, and the origin lies on both surfaces trivially.

   (c) Solve $\cos t=1+\sin t$ and watch for extraneous roots. On $C$, $x=y$ reads $\cos t=1+\sin t$. Squaring both sides,

   $${\cos }^{2}t=1+2\sin t+{\sin }^{2}t⟹1-{\sin }^{2}t=1+2\sin t+{\sin }^{2}t⟹0=2\sin t(1+\sin t).$$

   So $\sin t=0$ or $\sin t=-1$. Squaring may have added solutions, so test each candidate in the original equation. $\sin t=0$ gives $t=0$ or $t=\pi$: at $t=0$, $\cos 0=1=1+0$ ✓; at $t=\pi$, $\cos \pi =-1\ne 1$ ✗, extraneous (it solves $\cos t=-(1+\sin t)$ instead). $\sin t=-1$ gives $t=\frac{3\pi }{2}$: $\cos \frac{3\pi }{2}=0=1-1$ ✓.

   The curve meets the plane $x=y$ at exactly two points: $\mathbf{r}(0)=(1,1,2)$ and $\mathbf{r}\left(\frac{3\pi }{2}\right)=(0,0,0)$. Both have $x=y$, and both satisfy $z=x^2+y^2$ and $z=2y$: $2=1+1=2\cdot 1$ and $0=0=0$. (A closed curve crossing a plane generically does so an even number of times, which is a comforting check.)

8. 8.Proof · 10 min

   In this question $\mathbf{r}(t)$ is a vector function with as many derivatives as each part needs. Quote the differentiation rule you use at every step.

   1. (a)Prove that $\frac{d}{dt}[\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }\left)\right]=\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime \prime })$. Two of the terms that appear vanish; say exactly why each one does.
   2. (b)Prove that if $\mathbf{r}(t)\cdot {\mathbf{r}}^{\prime }(t)=0$ for every $t$, then $|\mathbf{r}(t)|$ is constant. (This is the converse of the statement that a constant-length vector function has $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=0$.)
   3. (c)Prove that wherever $\mathbf{r}(t)\ne 0$, $\frac{d}{dt}|\mathbf{r}(t)|=\frac{\mathbf{r}(t)\cdot {\mathbf{r}}^{\prime }(t)}{|\mathbf{r}(t)|}$. Then, for $\mathbf{r}(t)=⟨t,t^2,2⟩$, compute at $t=1$ both this rate of change and the speed $|{\mathbf{r}}^{\prime }(1)|$, and explain geometrically why they differ and which one must be the larger.
   Show the solution to question 8

   (a) Product rule for the dot, then for the cross. By the dot product rule,

   $$\frac{d}{dt}[\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })]={\mathbf{r}}^{\prime }\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })+\mathbf{r}\cdot \frac{d}{dt}({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }).$$

   The first term is zero because ${\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }$ is perpendicular to both of its factors, in particular to ${\mathbf{r}}^{\prime }$, and the dot product of perpendicular vectors is $0$. For the second, the cross product rule (factors kept in order) gives

   $$\frac{d}{dt}({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })={\mathbf{r}}^{\prime \prime }\times {\mathbf{r}}^{\prime \prime }+{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime \prime }.$$

   Here ${\mathbf{r}}^{\prime \prime }\times {\mathbf{r}}^{\prime \prime }=0$, because the cross product of a vector with itself has length $|{\mathbf{r}}^{\prime \prime }{|}^2\sin 0=0$ (a vector is parallel to itself). What survives is

   $$\frac{d}{dt}[\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })]=\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime \prime }).■$$

   (b) Differentiate the squared length. Let $f(t)=|\mathbf{r}(t){|}^2=\mathbf{r}(t)\cdot \mathbf{r}(t)$. By the dot product rule,

   $$f^{\prime }(t)={\mathbf{r}}^{\prime }\cdot \mathbf{r}+\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2\mathbf{r}(t)\cdot {\mathbf{r}}^{\prime }(t)=0$$

   for every $t$, by hypothesis. A real function with zero derivative on an interval is constant, so $|\mathbf{r}(t){|}^2=c$ for some constant $c\ge 0$, and therefore $|\mathbf{r}(t)|=\sqrt{c}$ is constant. $■$ (The hypothesis says the velocity is always tangent to the sphere through the current point; the conclusion is that the point never leaves that sphere.)

   (c) Chain rule on the square root. Where $\mathbf{r}\ne 0$ the function $|\mathbf{r}|=(\mathbf{r}\cdot \mathbf{r}{)}^{1/2}$ is a composition of the differentiable function $u\mapsto \sqrt{u}$ (for $u>0$) with $\mathbf{r}\cdot \mathbf{r}$, so by the chain rule and the dot product rule,

   $$\frac{d}{dt}|\mathbf{r}|=\frac{1}{2\sqrt{\mathbf{r}\cdot \mathbf{r}}}\cdot 2\mathbf{r}\cdot {\mathbf{r}}^{\prime }=\frac{\mathbf{r}\cdot {\mathbf{r}}^{\prime }}{|\mathbf{r}|}.■$$

   For $\mathbf{r}(t)=⟨t,t^2,2⟩$: ${\mathbf{r}}^{\prime }(t)=⟨1,2t,0⟩$. At $t=1$, $\mathbf{r}(1)=⟨1,1,2⟩$, ${\mathbf{r}}^{\prime }(1)=⟨1,2,0⟩$, $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=1+2+0=3$ and $|\mathbf{r}(1)|=\sqrt{6}$. So the distance from the origin is changing at the rate

   $$\frac{d}{dt}|\mathbf{r}|{|}_{t=1}=\frac{3}{\sqrt{6}}=\frac{\sqrt{6}}{2},$$

   while the speed is $|{\mathbf{r}}^{\prime }(1)|=\sqrt{1+4+0}=\sqrt{5}$.

   They differ because the formula just proved is the scalar projection of ${\mathbf{r}}^{\prime }$ onto $\mathbf{r}$: $\frac{\mathbf{r}\cdot {\mathbf{r}}^{\prime }}{|\mathbf{r}|}={\mathrm{comp}}_{\mathbf{r}}{\mathbf{r}}^{\prime }=|{\mathbf{r}}^{\prime }|\cos \varphi$, where $\varphi$ is the angle between the position and the velocity. Only the radial part of the velocity changes the distance from the origin; the part perpendicular to $\mathbf{r}$ moves the point sideways around the origin without changing how far away it is. Since $|\cos \varphi |\le 1$, the rate of change of distance can never exceed the speed, and indeed $\frac{\sqrt{6}}{2}<\sqrt{5}$ because $\frac{6}{4}<5$.

9. 9.Computation · 9 min

   Let ${\mathbf{r}}_1(t)=⟨t,1-t,3+t^2⟩$ and ${\mathbf{r}}_2(s)=⟨3-s,s-2,s^2⟩$ be two curves.

   1. (a)Evaluate ${\int }_0^1⟨\frac{1}{1+t^2},\frac{2t}{1+t^2},t{e}^t⟩dt$ exactly.
   2. (b)Show that the two curves meet at exactly one point, find it, and find the exact angle between their tangent vectors there.
   3. (c)Write parametric equations for the tangent line to ${\mathbf{r}}_1$ at the point found in (b), and find where that line crosses the $xy$-plane.
   4. (d)Show that both curves lie in one and the same plane, give its equation, and check that both tangent vectors from (b) are parallel to it.
   Show the solution to question 9

   (a) Integrate componentwise. Each component is an ordinary integral:

   $${\int }_0^1\frac{dt}{1+t^2}=[\arctan t{]}_0^1=\frac{\pi }{4},{\int }_0^1\frac{2tdt}{1+t^2}=[\ln (1+t^2){]}_0^1=\ln 2,$$

   and, integrating by parts with $u=t$, $dv=e^{t}dt$,

   $${\int }_0^{1}t{e}^{t}dt=[t{e}^t-e^t{]}_0^1=(e-e)-(0-1)=1.$$

   The integral is $⟨\frac{\pi }{4},\ln 2,1⟩$.

   (b) Match all three coordinates with independent parameters. A common point needs $t$ and $s$ with

   $$t=3-s,1-t=s-2,3+t^2=s^2.$$

   Substituting the first into the second gives $1-(3-s)=s-2$, which is $s-2=s-2$: an identity, so the first two equations impose only one condition, $t=3-s$. (Both curves project onto the same line $x+y=1$ in the $xy$-plane, which is why.) The third equation decides everything: $3+(3-s{)}^2=s^2$ gives $3+9-6s+s^2=s^2$, so $6s=12$, $s=2$ and $t=1$. There is exactly one solution, hence exactly one meeting point:

   $${\mathbf{r}}_1(1)=(1,0,4)={\mathbf{r}}_2(2).$$

   Tangent vectors: ${\mathbf{r}}_1^{\prime }(t)=⟨1,-1,2t⟩$ so ${\mathbf{r}}_1^{\prime }(1)=⟨1,-1,2⟩$; ${\mathbf{r}}_2^{\prime }(s)=⟨-1,1,2s⟩$ so ${\mathbf{r}}_2^{\prime }(2)=⟨-1,1,4⟩$. By the geometric form of the dot product,

   $$\cos \theta =\frac{⟨1,-1,2⟩\cdot ⟨-1,1,4⟩}{|⟨1,-1,2⟩||⟨-1,1,4⟩|}=\frac{-1-1+8}{\sqrt{6}\cdot \sqrt{18}}=\frac{6}{\sqrt{108}}=\frac{6}{6\sqrt{3}}=\frac{1}{\sqrt{3}},$$

   so $\theta =\arccos \frac{1}{\sqrt{3}}=\arccos \frac{\sqrt{3}}{3}$, an acute angle (the dot product is positive).

   (c) Point plus direction. The tangent line passes through $(1,0,4)$ with direction ${\mathbf{r}}_1^{\prime }(1)=⟨1,-1,2⟩$:

   $$x=1+u,y=-u,z=4+2u.$$

   It meets the $xy$-plane where $z=0$: $4+2u=0$, $u=-2$, giving the point $(-1,2,0)$.

   (d) Find a linear relation among the coordinates. On ${\mathbf{r}}_1$, $x+y=t+(1-t)=1$ for every $t$; on ${\mathbf{r}}_2$, $x+y=(3-s)+(s-2)=1$ for every $s$. So both curves lie in the plane $x+y=1$, whose normal is $\mathbf{n}=⟨1,1,0⟩$. A vector is parallel to the plane exactly when it is perpendicular to $\mathbf{n}$: $⟨1,-1,2⟩\cdot \mathbf{n}=1-1+0=0$ and $⟨-1,1,4⟩\cdot \mathbf{n}=-1+1+0=0$, as they must be, since the tangent to a curve lying in a plane lies in that plane. (Within the plane $x+y=1$ each curve is a parabola in the height $z$; two parabolas of the same width $z\sim t^2$, shifted, cross exactly once, which is the geometric reason behind (b).)

10. 10.Computation · 12 min

    Let $\mathbf{r}(t)=⟨e^t\cos t,e^t\sin t,e^t⟩$ for $t\in \mathbb{R}$.

    1. (a)Show that the curve lies on the cone $z^2=x^2+y^2$, and say on which part of it.
    2. (b)Find $|{\mathbf{r}}^{\prime }(t)|$ and show it simplifies to a single exponential. Find the length of the curve for $0\le t\le \ln 2$, the arc length function $s(t)$ measured from $t=0$, and $t$ as a function of $s$.
    3. (c)Compute $\kappa (t)$ using $\kappa =\frac{|{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }|}{|{\mathbf{r}}^{\prime }{|}^3}$, simplify fully, and state $\kappa (0)$. What happens to $\kappa$ as $t\to \infty$, and what does that say about the shape of the curve?
    4. (d)Show that the angle between the tangent vector ${\mathbf{r}}^{\prime }(t)$ and the $z$-axis is the same at every point of the curve, and find its cosine.
    Show the solution to question 10

    (a) Check the equation of the cone. $x^2+y^2=e^{2t}{\cos }^{2}t+e^{2t}{\sin }^{2}t=e^{2t}=z^2$, so every point of the curve satisfies $z^2=x^2+y^2$. Since $z=e^t>0$ the curve lies on the upper nappe $z=\sqrt{x^2+y^2}$, spiralling outward and upward as $t$ increases (the distance from the $z$-axis is $\sqrt{x^2+y^2}=e^t$, which grows, while the angle in the $xy$-plane is $t$).

    (b) Speed, length, arc length function. Differentiate each component with the product rule:

    $${\mathbf{r}}^{\prime }(t)=⟨e^t(\cos t-\sin t),e^t(\sin t+\cos t),e^t⟩.$$

    Squaring and adding, the mixed terms cancel:

    $$|{\mathbf{r}}^{\prime }(t){|}^2=e^{2t}[(\cos t-\sin t{)}^2+(\sin t+\cos t{)}^2+1]=e^{2t}[(1-2\sin t\cos t)+(1+2\sin t\cos t)+1]=3e^{2t},$$

    so $|{\mathbf{r}}^{\prime }(t)|=\sqrt{3}{e}^t$. The speed is never zero and ${\mathbf{r}}^{\prime }$ is continuous, so the curve is smooth and the arc length formula applies:

    $$L={\int }_0^{\ln 2}\sqrt{3}{e}^{t}dt=\sqrt{3}[e^t{]}_0^{\ln 2}=\sqrt{3}(2-1)=\sqrt{3}.$$

    The arc length function from $t=0$ is $s(t)={\int }_0^t\sqrt{3}{e}^{u}du=\sqrt{3}(e^t-1)$. Solving, $e^t=1+\frac{s}{\sqrt{3}}$, so $t=\ln \left(1+\frac{s}{\sqrt{3}}\right)$ for $s\ge 0$. (Check: $s=\sqrt{3}$ gives $t=\ln 2$, matching $L$.)

    (c) Curvature by the cross product formula. Differentiate ${\mathbf{r}}^{\prime }$ once more:

    $${\mathbf{r}}^{\prime \prime }(t)=⟨e^t(\cos t-\sin t)+e^t(-\sin t-\cos t),e^t(\sin t+\cos t)+e^t(\cos t-\sin t),e^t⟩=⟨-2e^t\sin t,2e^t\cos t,e^t⟩.$$

    Factor $e^t$ out of each vector and write $c=\cos t$, $s=\sin t$ for brevity: ${\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }=e^{2t}⟨c-s,s+c,1⟩\times ⟨-2s,2c,1⟩$. Expanding the determinant,

    $$⟨c-s,s+c,1⟩\times ⟨-2s,2c,1⟩=⟨(s+c)-2c,-[(c-s)+2s],2c(c-s)+2s(s+c)⟩=⟨s-c,-s-c,2⟩,$$

    since $2c^2-2cs+2s^2+2cs=2$. Its squared length is $(s-c{)}^2+(s+c{)}^2+4=2+4=6$, so $|{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }|=\sqrt{6}{e}^{2t}$. (Check that $⟨s-c,-s-c,2⟩$ is perpendicular to $⟨c-s,s+c,1⟩$: $-(s-c{)}^2-(s+c{)}^2+2=-2+2=0$.) Hence

    $$\kappa (t)=\frac{\sqrt{6}{e}^{2t}}{(\sqrt{3}{e}^t{)}^3}=\frac{\sqrt{6}}{3\sqrt{3}}{e}^{-t}=\frac{\sqrt{2}}{3}{e}^{-t},\kappa (0)=\frac{\sqrt{2}}{3}.$$

    As $t\to \infty$, $\kappa \to 0$: the spiral's turns get ever larger in radius (the distance from the axis is $e^t$), and a bend of larger radius is gentler. The curvature falls off like the reciprocal of that radius, exactly as for circles.

    (d) Dot the tangent with $\mathbf{k}$. By the geometric form of the dot product the angle $\varphi$ between ${\mathbf{r}}^{\prime }(t)$ and $\mathbf{k}=⟨0,0,1⟩$ satisfies

    $$\cos \varphi =\frac{{\mathbf{r}}^{\prime }(t)\cdot \mathbf{k}}{|{\mathbf{r}}^{\prime }(t)||\mathbf{k}|}=\frac{e^t}{\sqrt{3}{e}^t}=\frac{1}{\sqrt{3}},$$

    independent of $t$. So the tangent makes the same angle $\arccos \frac{1}{\sqrt{3}}$ with the axis of the cone at every point: the curve climbs the cone at a constant pitch, the way a helix climbs a cylinder. (The same computation with $\mathbf{r}$ in place of $\mathbf{k}$ gives $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2e^{2t}$ and $|\mathbf{r}|=\sqrt{2}{e}^t$, so the tangent also makes a constant angle with the position vector, with cosine $\sqrt{2/3}$.)

11. 11.Critique · 9 min

    A particle moves with position $\mathbf{r}(t)=⟨3\cos 2t,3\sin 2t,4t⟩$. A student is asked for the tangential and normal components of its acceleration and submits the following.

    Step 1. $\mathbf{v}(t)=⟨-6\sin 2t,6\cos 2t,4⟩$, so the speed is $v=\sqrt{36+16}=2\sqrt{13}$, a constant.

    Step 2. Since the speed is constant, $a_T=\frac{dv}{dt}=0$.

    Step 3. $\mathbf{T}=\frac{\mathbf{v}}{2\sqrt{13}}$, so ${\mathbf{T}}^{\prime }(t)=\frac{1}{2\sqrt{13}}⟨-12\cos 2t,-12\sin 2t,0⟩$ and $|{\mathbf{T}}^{\prime }(t)|=\frac{12}{2\sqrt{13}}=\frac{6}{\sqrt{13}}$. Therefore the curvature is $\kappa =|{\mathbf{T}}^{\prime }(t)|=\frac{6}{\sqrt{13}}$.

    Step 4. Hence $a_N=\kappa v^2=\frac{6}{\sqrt{13}}\cdot 52=24\sqrt{13}$.

    Step 5. As a check, $\mathbf{a}(t)=⟨-12\cos 2t,-12\sin 2t,0⟩$ has $|\mathbf{a}|=12$. This is not equal to $\sqrt{a_T^2+a_N^2}=24\sqrt{13}$, so the identity $|\mathbf{a}{|}^2=a_T^2+a_N^2$ must require some extra hypothesis that this motion does not satisfy.

    1. (a)Identify every step that is wrong. For each, say precisely what has been misapplied and why.
    2. (b)Give the correct curvature and the correct $a_N$, and show that the check in Step 5 now passes.
    3. (c)Confirm $\kappa$ and $a_N$ independently by the cross product formulas $\kappa =\frac{|{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }|}{|{\mathbf{r}}^{\prime }{|}^3}$ and $a_N=\frac{|{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }|}{|{\mathbf{r}}^{\prime }|}$. Then state the radius of the osculating circle and explain why it is larger than the radius $3$ of the cylinder the path winds around.
    Show the solution to question 11

    (a) Steps 1 and 2 are correct; Steps 3 and 5 are wrong; Step 4 inherits the error of Step 3.

    Step 3 confuses turning per unit time with turning per unit arc length. Curvature is defined as $\kappa =\left|\frac{d\mathbf{T}}{ds}\right|$, and converting to the parameter $t$ costs a factor of the speed: $\kappa =\frac{|{\mathbf{T}}^{\prime }(t)|}{|{\mathbf{r}}^{\prime }(t)|}$. The formula $\kappa =|{\mathbf{T}}^{\prime }(t)|$ is valid only when $t$ is arc length, i.e. when the speed is $1$; here the speed is $2\sqrt{13}$. The computation of ${\mathbf{T}}^{\prime }$ itself is right (differentiating $\mathbf{v}/(2\sqrt{13})$ is legitimate precisely because the speed is constant), but the last equality of the step is not.

    Step 5 draws the wrong conclusion from a failed check. The identity $|\mathbf{a}{|}^2=a_T^2+a_N^2$ has no extra hypothesis: it holds for every smooth motion, because $\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}$ with $\mathbf{T}$ and $\mathbf{N}$ perpendicular unit vectors, and Pythagoras does the rest. A check that fails is evidence of an error upstream, never evidence against the theorem. The student computed $|\mathbf{a}|=12$ correctly and should have gone back to look for the mistake.

    (b) Divide by the speed. With $|{\mathbf{T}}^{\prime }|=\frac{6}{\sqrt{13}}$ and $|{\mathbf{r}}^{\prime }|=2\sqrt{13}$,

    $$\kappa =\frac{|{\mathbf{T}}^{\prime }(t)|}{|{\mathbf{r}}^{\prime }(t)|}=\frac{6/\sqrt{13}}{2\sqrt{13}}=\frac{6}{26}=\frac{3}{13},$$

    a constant, as it should be for a helix. Then

    $$a_N=\kappa v^2=\frac{3}{13}\cdot 52=12.$$

    The check now passes: $a_T^2+a_N^2=0+144=144=|\mathbf{a}{|}^2$. With $a_T=0$ the whole acceleration is normal, and indeed $\mathbf{a}=⟨-12\cos 2t,-12\sin 2t,0⟩$ points horizontally toward the axis, which is the direction of $\mathbf{N}$.

    (c) The cross product route. With ${\mathbf{r}}^{\prime }=⟨-6\sin 2t,6\cos 2t,4⟩$ and ${\mathbf{r}}^{\prime \prime }=⟨-12\cos 2t,-12\sin 2t,0⟩$,

    $${\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }=⟨6\cos 2t\cdot 0-4(-12\sin 2t),-[(-6\sin 2t)\cdot 0-4(-12\cos 2t)],(-6\sin 2t)(-12\sin 2t)-(6\cos 2t)(-12\cos 2t)⟩$$

    $$=⟨48\sin 2t,-48\cos 2t,72{\sin }^{2}2t+72{\cos }^{2}2t⟩=⟨48\sin 2t,-48\cos 2t,72⟩.$$

    Its length is $\sqrt{48^2+72^2}=\sqrt{2304+5184}=\sqrt{7488}=\sqrt{576\cdot 13}=24\sqrt{13}$. Therefore

    $$\kappa =\frac{24\sqrt{13}}{(2\sqrt{13}{)}^3}=\frac{24\sqrt{13}}{8\cdot 13\sqrt{13}}=\frac{24}{104}=\frac{3}{13},a_N=\frac{24\sqrt{13}}{2\sqrt{13}}=12,$$

    agreeing with (b). (Notice that the student's wrong $a_N=24\sqrt{13}$ is exactly $|{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }|$ with the division by $|{\mathbf{r}}^{\prime }|$ omitted — the same missing factor in a different guise.)

    The osculating circle has radius $\rho =\frac{1}{\kappa }=\frac{13}{3}$, which exceeds the cylinder's radius $3$. The horizontal projection of the path is a circle of radius $3$, but the actual path also climbs: the vertical velocity $4$ stretches each turn into a longer, gentler arc, so the curve bends less sharply than its shadow does. Only if the climb were removed ($z$ constant) would $\kappa$ rise to $\frac{1}{3}$ and $\rho$ fall to $3$.

12. 12.Proof · 12 min

    Let $\omega >0$ be a constant. A particle has position $\mathbf{r}(t)$, velocity $\mathbf{v}={\mathbf{r}}^{\prime }$ and acceleration $\mathbf{a}={\mathbf{r}}^{\prime \prime }$, and the acceleration always points toward the origin in proportion to the displacement:

    $$\mathbf{a}(t)=-{\omega }^2\mathbf{r}(t)\text{for all }t.$$

    Name the differentiation rule and the vector identity used at every step of (a) and (b).

    1. (a)Prove that $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ is a constant vector, and deduce that if $\mathbf{h}\ne 0$ the particle moves in a fixed plane through the origin. Identify that plane.
    2. (b)Prove that $E=|\mathbf{v}{|}^2+{\omega }^2|\mathbf{r}{|}^2$ is constant in time.
    3. (c)Verify that $\mathbf{r}(t)=⟨\cos \omega t,2\sin \omega t,0⟩$ satisfies $\mathbf{a}=-{\omega }^2\mathbf{r}$, and compute $\mathbf{h}$ and $E$ for it, confirming that both are constant.
    4. (d)For the motion in (c), find $a_T$, $a_N$ and the curvature $\kappa$ at $t=0$, and the radius of the osculating circle there. The speed of this motion is not constant; explain why $a_T=0$ at $t=0$ nonetheless.
    Show the solution to question 12

    (a) Differentiate the cross product. By the cross product rule (factors kept in order),

    $$\frac{d\mathbf{h}}{dt}=\frac{d}{dt}(\mathbf{r}\times \mathbf{v})={\mathbf{r}}^{\prime }\times \mathbf{v}+\mathbf{r}\times {\mathbf{v}}^{\prime }=\mathbf{v}\times \mathbf{v}+\mathbf{r}\times \mathbf{a}.$$

    The first term is $0$ because any vector is parallel to itself (sine of the angle is $0$). For the second, substitute the hypothesis and pull the scalar out: $\mathbf{r}\times \mathbf{a}=\mathbf{r}\times (-{\omega }^2\mathbf{r})=-{\omega }^2(\mathbf{r}\times \mathbf{r})=0$, again because parallel vectors have zero cross product. So ${\mathbf{h}}^{\prime }(t)=0$ for all $t$; each component has zero derivative, hence is constant, and $\mathbf{h}$ is a constant vector. $■$

    Now $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ is perpendicular to both of its factors, so $\mathbf{r}(t)\cdot \mathbf{h}=0$ for every $t$. If $\mathbf{h}\ne 0$ this says the position vector always satisfies the single linear equation $\mathbf{h}\cdot \mathbf{r}=0$: the particle stays in the plane through the origin with normal vector $\mathbf{h}$. It is the plane spanned by the initial position and initial velocity, $\mathbf{r}(0)$ and $\mathbf{v}(0)$, since $\mathbf{h}=\mathbf{r}(0)\times \mathbf{v}(0)$.

    (b) Differentiate the two squared lengths. Using $|\mathbf{u}{|}^2=\mathbf{u}\cdot \mathbf{u}$ and the dot product rule,

    $$\frac{dE}{dt}=2\mathbf{v}\cdot {\mathbf{v}}^{\prime }+{\omega }^2\cdot 2\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2\mathbf{v}\cdot \mathbf{a}+2{\omega }^2\mathbf{r}\cdot \mathbf{v}.$$

    Substitute $\mathbf{a}=-{\omega }^2\mathbf{r}$ in the first term and use commutativity of the dot product:

    $$\frac{dE}{dt}=-2{\omega }^2\mathbf{v}\cdot \mathbf{r}+2{\omega }^2\mathbf{r}\cdot \mathbf{v}=0.$$

    A real function with zero derivative is constant, so $E$ is conserved. $■$ (Physically $E$ is twice the energy per unit mass of a particle on an ideal spring: kinetic plus potential.)

    (c) Check the example. $\mathbf{v}(t)=⟨-\omega \sin \omega t,2\omega \cos \omega t,0⟩$ and $\mathbf{a}(t)=⟨-{\omega }^2\cos \omega t,-2{\omega }^2\sin \omega t,0⟩=-{\omega }^2⟨\cos \omega t,2\sin \omega t,0⟩=-{\omega }^2\mathbf{r}(t)$. ✓

    Since both $\mathbf{r}$ and $\mathbf{v}$ have zero third component, $\mathbf{r}\times \mathbf{v}$ has only a $\mathbf{k}$-component, equal to $r_1{v}_2-r_2{v}_1$:

    $$\mathbf{h}=⟨0,0,\cos \omega t\cdot 2\omega \cos \omega t-2\sin \omega t\cdot (-\omega \sin \omega t)⟩=⟨0,0,2\omega ⟩,$$

    constant, and the plane $\mathbf{h}\cdot \mathbf{r}=0$ is $z=0$, where the motion indeed takes place. For the energy,

    $$E=({\omega }^2{\sin }^2\omega t+4{\omega }^2{\cos }^2\omega t)+{\omega }^2({\cos }^2\omega t+4{\sin }^2\omega t)={\omega }^2(5{\sin }^2\omega t+5{\cos }^2\omega t)=5{\omega }^2,$$

    constant. (The path is the ellipse $x^2+\frac{y^2}{4}=1$, traversed once per period $2\pi /\omega$.)

    (d) Components at $t=0$. ${\mathbf{r}}^{\prime }(0)=⟨0,2\omega ,0⟩$, so the speed is $v(0)=2\omega$, and ${\mathbf{r}}^{\prime \prime }(0)=⟨-{\omega }^2,0,0⟩$. Then

    $$a_T=\frac{{\mathbf{r}}^{\prime }\cdot {\mathbf{r}}^{\prime \prime }}{|{\mathbf{r}}^{\prime }|}=\frac{0}{2\omega }=0,{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }=⟨0,0,0-2\omega \cdot (-{\omega }^2)⟩=⟨0,0,2{\omega }^3⟩,a_N=\frac{2{\omega }^3}{2\omega }={\omega }^2.$$

    Check: $|\mathbf{a}(0)|={\omega }^2=\sqrt{0+{\omega }^4}$. ✓ The curvature is $\kappa =\frac{a_N}{v^2}=\frac{{\omega }^2}{4{\omega }^2}=\frac{1}{4}$ (equivalently $\frac{|{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }|}{|{\mathbf{r}}^{\prime }{|}^3}=\frac{2{\omega }^3}{8{\omega }^3}$), so the osculating circle at $(1,0,0)$ has radius $\rho =4$. The value is independent of $\omega$, as it must be: curvature is a property of the ellipse, not of how fast it is traversed.

    Why $a_T=0$ there: $a_T$ is the derivative of the speed, not the speed itself. Here $v(t{)}^2={\omega }^2{\sin }^2\omega t+4{\omega }^2{\cos }^2\omega t={\omega }^2(1+3{\cos }^2\omega t)$, which is not constant — it oscillates between $\omega$ and $2\omega$ — but at $t=0$ it is at its maximum $2\omega$, and a differentiable function has zero derivative at an interior maximum. So the speed is momentarily not changing at $t=0$, even though it is about to decrease; all of the acceleration at that instant goes into turning. (At $t=\frac{\pi }{2\omega }$, the end of the long axis, the same computation gives $a_T=0$ again, $a_N=2{\omega }^2$, $v=\omega$ and $\kappa =2$: the ellipse bends most sharply at the ends of its major axis and least at the ends of its minor axis.)